Question

Let $n\ge 5$ and $b\ne 0$. In the binomial expansion of ${(a-b)}^n$, the sum of the 5th and 6th terms is zero then $a/b$ equals

• A. $\frac{5}{n-4}$
• B. $\frac{1}{5(n-4)}$
• C. $\frac{n-5}{6}$
• D. $\frac{n-4}{5}$

Answer 1

The correct option is A $\frac{5}{n-4}$

Solution:

Given that:

$n\ge 5,b\ne 0,$ given expansion is $(a-b{)}^n$ and

$t_5+t_6=0$

To find: $\frac{a}{b}=?$

Solution:

$t_5={}^n{C}_4{a}^4(-b{)}^{(n-4)}$ and

$t_6={}^n{C}_5{a}^5(-b{)}^{(n-5)}$

$\because t_5+t_6=0$

$\therefore {}^n{C}_4{a}^4(-b{)}^{(n-4)}$$+{}^n{C}_5{a}^5(-b{)}^{(n-5)}=0$

$⟹{}^n{C}_{5}a={}^n{C}_{4}b$

$⟹\frac{a}{b}=\frac{{}^n{C}_4}{{}^5{C}_5}=\frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}}=\frac{5}{n-4}$

Hence, A is the correct answer.